\section{Exterior Derivative}

With $n$ grassmann generators, we can write
\begin{equation}
d = \sum_{i}\varepsilon^{i}\frac{\partial}{\partial x^i}
\end{equation}
where in cartesian coordinates $x^1=x$, $x^2=y$, $x^3=z$, etc. Observe that
by direct computation we find
\begin{equation}
d^2 = \left(\sum_{i}\varepsilon^{i}\frac{\partial}{\partial x^i} \right)^2.
\end{equation}
We will perform a proof by induction that
\begin{equation}
d^2 = 0.
\end{equation}
We will use the notation
\begin{equation}
\partial_{i} = \frac{\partial}{\partial x^i}.
\end{equation}

\begin{proof}
\noindent\textbf{Base Case} We see that with
\begin{subequations}
\begin{align}
\left(\varepsilon^1\partial_1 + \varepsilon^2\partial_2 \right)^2 &= (\varepsilon^1\partial_1)^2 + (\varepsilon^2\partial_2)^2 + (\varepsilon^1\varepsilon^2\partial_1\partial_2) + (\varepsilon^2\varepsilon^1\partial_1\partial_2)\\
&= (0) + (0) + (\varepsilon^1\varepsilon^2+\varepsilon^2\varepsilon^1)\partial_1\partial_2 \\
&= (0)\partial_1\partial_2\\
&= 0.
\end{align}
\end{subequations}
So it is true with $n=2$.

\noindent\textbf{Inductive Hypothesis} We assume that it works with $n$.

\noindent\textbf{Inductive Case} With $n+1$ we see that
\begin{equation}
\left(\varepsilon^1\partial_1 +\ldots+\varepsilon^n\partial_n+ \varepsilon^{n+1}\partial_{n+1} \right)^2 = (z + \varepsilon^{n+1}\partial_{n+1})^2
\end{equation}
where we define
\begin{equation}
z = \varepsilon^1\partial_1 +\ldots+\varepsilon^n\partial_n.
\end{equation}
By the inductive hypothesis, we assumed that
\begin{equation}
z^2 = 0
\end{equation}
so we find
\begin{equation}
(z + \varepsilon^{n+1}\partial_{n+1})^2 = \cancelto{0}{z^2} + \cancelto{0}{(\varepsilon^{n+1}\partial_{n+1})^2} + z\varepsilon^{n+1}\partial_{n+1} + \varepsilon^{n+1}\partial_{n+1}z.
\end{equation}
So we have a simple expression, and it's one we all know and love! It's simply
\begin{equation}
z\varepsilon^{n+1}\partial_{n+1} + \varepsilon^{n+1}\partial_{n+1}z = \left(\sum_{i=1}^{n}\varepsilon{i}\partial_{i}\right)\varepsilon^{n+1}\partial_{n+1} + \varepsilon^{n+1}\partial_{n+1}\left(\sum_{i=1}^{n}\varepsilon{i}\partial_{i}\right)
\end{equation}
but by our expression (\ref{generalizationOfCliffordAlgebras}), we see that
these two terms cancel each other out! That is, we get a number of expressions
of the form
\begin{equation}
\sum_{i=1}^{n}\varepsilon^{n+1}\varepsilon^{i} + \varepsilon^{i}\varepsilon^{n+1} = \sum_{i=1}^{n}(0) = 0.
\end{equation}
This completes our proof by induction!
\end{proof}
